# Limits and convolution

Let $f,g \in L^2(\mathbb{R^n})$, $\{ f_n \}, \{ g_m \} \subset C^\infty_0(\mathbb{R}^n)$ (infinitely differentiable functions with compact support) where $f_n \to f$ in $L^2$, and $g_n \to g$ in $L^2$.

Then $f_n \star g_m(x) \to f \star g(x)$ pointwise as $n,m \to \infty$ (where $\star$ denotes convolution).

Is it true that $\int_{\mathbb{R}^n} f_n \star g_m dx \to \int_{\mathbb{R}^n} f \star g dx$?

I've thought about using the dominated convergence theorem, but I couldn't figure out how, since I don't really know what I can do with the sequences of functions given only the facts at hand.

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No it is not true.

First note that we can improve pointwise convergence to uniform convergence of $f_{n} \ast g_{m} \to f \ast g$. This is an easy consequence of Hölder's inequality (markup doesn't like my subscripts, so I'll use $\|\cdot\|$ for the $L^{2}$-norm): $|(f \ast g)(x)| \leq \int |f(x - y) g(y)|\,dy \leq \|f\| \|g\|.$ As $f_{n} \ast g_{m} - f \ast g = f_{n} \ast (g_{m} - g) + (f_{n} - f) \ast g$ we get $|(f_{n} \ast g_{m})(x) - (f \ast g)(x)| \leq \|f_{n}\| \|g_{m} - g\| + \|f_{n} - f\| \|g\|.$ Since $\|f_{n}\| \to \|f\|$ and both $\|f_{n} - f\| \to 0$ and $\|g_{m} - g\| \to 0$, the right hand side can be made arbitrarily small independently of $x$. It follows in particular that $f \ast g$ is continuous and vanishes at infinity (that's one reason why I prefer writing $C_{c}$ for functions with compact support and $C_{0}$ for functions vanishing at infinity).

But we cannot do better, i.e., get $L^1$-convergence without further hypotheses. To see what happens, I prefer to ignore the condition that $f_{n} \in C^{\infty}$ but rather look at the function $f_{n} = \frac{1}{n}[-n,n]$. Then $f_{n} \to 0$ in $L^{2}$ but not in $L^{1}$. By the above argument we have $f_{n} \ast f_{n} \to 0$ uniformly on $\mathbb{R}$. On the other hand it is easy to see that for $x \in [-\frac{n}{2},\frac{n}{2}]$ $(f_{n} \ast f_{n})(x) = \int f_{n}(y) f_{n}(x-y)\,dy \geq \frac{1}{n^{2}} \frac{n}{2} = \frac{1}{2n}$ (the intervals $[-n,n]$ and $[-n-x,n-x]$ have an overlap of length at least $\frac{n}{2}$) and therefore we have the estimate $\int (f_{n} \ast f_{n}) \geq \frac{1}{2}$. Thus we cannot have $\int f_{n} \ast f_{n} \to \int f \ast f = 0$. It is now easy to cook up an example along these lines with smooth functions.

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\lVert \cdot \rVert_2 works: $\lVert \cdot \rVert_2$ – kahen Jan 30 '11 at 17:20
@kahen: Thanks, I'll try it next time. I usually write \| \cdot \|_2 and this is getting mangled up after two or three instnaces. – t.b. Jan 30 '11 at 20:06

For any integrable $f$ and $g$ (so in particular for functions in $C_O({\mathbb R}^n)$ one has $\int_{{\mathbb R}^n} f \ast g = \int_{{\mathbb R}^n} f \int_{{\mathbb R}^n} g$. So your question is asking if $\int_{{\mathbb R}^n} f_n \int_{{\mathbb R}^n} g_n$ necessarily converges to $\int_{{\mathbb R}^n} f \int_{{\mathbb R}^n} g$. A counterexample will be provided by letting $f_n = g_n$ such that $f_n$ converges to $f$ in $L^2$ but such that $\int_{{\mathbb R}^n} f_n$ does not converge to $\int_{{\mathbb R}^n} f$. To do this, one can let $\phi(x)$ be any smooth function with compact support and integral one, and then let $f_n(x) = {1 \over n} \phi({x \over n})$. Each $f_n$ will have integral $1$, but $f_n(x)$ converges in $L^2$ to $f(x) = 0$ which has integral zero.

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That's a much nicer way of putting it than mine... My example is just obtained by this procedure by taking $\phi = [-1,1]$ as a starting point. One still needs to argue why $f_{n} \to 0$ in $L^2$ but that's very easy. – t.b. Jan 30 '11 at 15:16
For some reason I thought he wanted pointwise rather than $L^2$ convergence.. anyhow I edited my answer accordingly. – Zarrax Jan 30 '11 at 15:33